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Let $T$ be normal. By the spectral theorem, there is a unitary map $V : H \to L^2(\mu)$ for a suitable measure $\mu$ on a measure space $X$ such that we have $$T = V^\ast M_f V$$ for some bounded function $f : X \to \Bbb{C}$, where $M_f : L^2(\mu) \to L^2 (\mu), g \mapsto f\cdot g$ is the multiplication operator which multiplies by $f$. By the usual ...

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So we want to show that if $A-\lambda I$ is not invertible, then $\lambda\geq0$. There are three ways in which $A_\lambda I$ may fail to be invertible: $\ker(A-\lambda I)\ne\{0\}$. In this case $\lambda$ is an eigenvalue. So there exists a unit vetor $v\in H$ with $Av=\lambda v$. Then $$\lambda=\langle\lambda v,v\rangle=\langle Av,v\rangle\geq0.$$ ...

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No because in a Banach space weak topology has the same bounded sets as the topology induced by the norm. This relies on the Banach–Steinhaus theorem. Certainly weakly convergent sequences are bounded in the weak topology. Let $X$ be a Banach space. Suppose that $A\subset X$ is bounded in the weak topology. Then $A$ regarded as a subset of $X^{**}$ is ...

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These hints can help you, provided that you know the definitions: First try to find what the limit $T$ is in the weak topology. By the Riesz representation theorem, any linear functional in $H$ is written as $\ell(x) = \langle x, \xi\rangle$ for some vector $\xi$. What happens with $\ell(T_n(x))$ when $n\rightarrow \infty$? Now consider $x = e_1$. What ...

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Let $V := \langle E\rangle$ be the span of $E$, i.e. the set of finite linear combinations of the orthonormal basis. It is easy to see $\langle x_n, v\rangle \to 0$. Now, let $y \in H$ be arbitrary and $\epsilon >0$. There is $v \in V$ with $\Vert v -y\Vert <\epsilon$, since $V$ is dense because $E$ is an orthonormal basis. Now, $$|\langle ... 3 Let ||x_n||<A Take M to be the subspace of finite linear combination of basis elements. Then it is easy to see that M is dense in H. \langle x_n,e\rangle \rightarrow 0 \forall e\implies \langle x_n,m\rangle \rightarrow 0 \forall m\in M Now fix y\in H. Take \epsilon >0 Choose m\in M such that ||m-y||<\epsilon A^{-1}  As ... 1 The answer to the first question is negative. The separable Hilbert space l^2 contains an uncountable linearly independent set. For example, any Hamel basis (over \mathbb C, or over \mathbb R if you prefer real Hilbert spaces) is uncountable. For a more explicit example, recall first that \mathbb N has an uncountable family of infinite subsets A_r, ... 2 Yes, T_n converges strongly to zero by the Riemann-Lebesgue lemma: For any x\in H,$$ \lim_{n\to \infty} \langle x,e_n\rangle = 0 $$which itself follows from Bessel's inequality$$ \sum_{n=1}^{\infty} |\langle x,e_n\rangle | ^2 \leq \|x\|^2 $$Furthermore, T_n does not converge in norm, because if n\neq m$$ \|T_n - T_m\| \geq \|T_n(e_n) - ...

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Part 2 is easier to answer: no, the way monotone operators are defined in functional analysis, $-F$ is not in general monotone when $F$ is. (This is unlike the concept of monotonicity in real analysis). Monotone operators correspond to (non-strictly) increasing functions. The reverse inequality defines dissipative operators. Part 1, geometric ...

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The answer to the question in the title is no. Let $H=\ell^2(\mathbb N)$, $$N=\{x\in H:\ \exists m:\ x(n)=0,\ \forall n\geq m\}$$ and $$M=\{\lambda z:\ \lambda\in\mathbb C\},$$ where $$z=\left(1,\frac12,\frac13,\frac14,\ldots\right).$$ Then $N$ is dense, $M$ is closed, and $N\cap M=\{0\}$.

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As you have defined: $$p_nf = \sum_{k=1}^{n}\left[n\int_{\frac{k-1}{n}}^{\frac{k}{n}}f(y)dy\right]\chi_{[\frac{k-1}{n},\frac{k}{n}]}(x)$$ The linear operator $p_n$ is an orthogonal projection operator onto the linear span of the orthonormal set $\left\{\sqrt{n}\chi_{[\frac{k-1}{n},\frac{k}{n}]}\right\}_{k=1}^{n}$. Hence, $p_n^2=p_n$ and $\|p_nf\| \le ... 2 Let$e_k = (0,0,0,\ldots,0,0,1,0,0,\ldots)$where the$1$is in the$k$th place and all other entries are$0$. This belongs to your proposed space. The set$\{e_k : k=1,2,3,\ldots\}$is linearly independent. 0 Define$T\colon M\to M'$by$(Tf)(x)=e^{-x^2/2}f(x)$. ClearThen, for amy$f,g\in M$we have $$\langle f,g\rangle=\int_{\mathbb R}f(x)\,g(x)\,e^{-x^2}\,dx=\int_{\mathbb R}f(x)\,e^{-x^2/2}\,g(x)\,e^{-x^2/2}\,dx=\langle Tf,Tg\rangle.$$ Since clearly$T$is bijective,$T$is an isometry and$M$and$M'$are isometric. -1 If your problem comes from a more strongest formulation like, for example, this ODE$-u''(x)+\mu u(x)=f(x)$with some boundary conditions try to use the condition of eigenvalue there. Continuing with the example, you know that$f \neq 0$will be an eigenvector with eigenvalue$\lambda$if and only if$-\lambda f''(x)+\mu \lambda f(x)=\lambda f(x)$. So you ... 1 First remember that$\{|u_1\rangle, |u_2\rangle, |u_3\rangle\} \not\subseteq \operatorname{span}(|\phi_1\rangle, |\phi_2\rangle)$-- in fact the span might not include any of them. So just start checking whether each of your basis vectors is in the span and stop when you find one that isn't. Clearly$|u_1\rangle \not\in \operatorname{span}(|\phi_1\rangle, ...

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Let us define the partial sum $$z_n = \sum_{k = 1}^n x_k.$$ Then, we have for $m > n$ (by orthogonality) $$\|z_m - z_n\|^2 = \left\|\sum_{k = n+1}^m x_k\right\|^2 = \sum_{k = n+1} \|x_k\|^2.$$ Hence, $\{z_n\}$ is a Cauchy sequence in $H$ iff $\{\|x_k\|^2\}$ is summable. Hence, it remains to show that $\{\|x_k\|^2\}$ is summable. Suppose that ...

2

This operator is known as the Volterra operator. As Martin R pointed out, its kernel is trivial. The range can't be described in simpler terms than "antiderivatives of $L^2$ functions", which is a tautology. The operator is compact. One way to show it is to apply a general theorem saying that all Hilbert-Schmidt operators are compact (as is gone here). ...

0

It is not enough to just know that $\dim U=\dim V$. For instance, suppose $U=H$ and $V$ is the orthogonal complement of a single nonzero vector $v\in H$. As long as $H$ is infinite-dimensional, $U$ and $V$ will have the same dimension, but clearly no unitary $T:H\to H$ can map $U$ to $V$, since then $T$ would not be surjective. More generally, one can see ...

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You need exactly $\dim U=\dim V$ and $\dim U^\perp=\dim V^\perp$, in the sense that they have bases with the same cardinality. Suppose that $\{e_j\}_{j\in J}$ is an orthonormal basis of $U$ and that $\{f_j\}_{j\in J}$ is an orthonormal basis of $V$. Extend both basis to orthonormal bases of $H$, $\{e_j\}_{j\in K}$, $\{f_j\}_{j\in K}$, where $K\supset ... 0 Could you explain me how we deduce that$d(x,K)=1$?$K$is the interior of the unit circle about the origin. If$y \in K$, then$d(\mathbf 0,y) < 1$. Now for$x = (0,2), d(\mathbf 0, x) = 2$. By the triangle inequality$d(\mathbf 0, x) \le d(\mathbf 0,y) + d(x, y)$. So $$d(x, y) \ge d(\mathbf 0, x) - d(\mathbf 0,y) > 2 - 1 = 1.$$ Therefore$d(x, ...

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A small remark. Note that both series converge naturally in $L^2$, considering that the basis functions are elements of the Hilbert space (and thus, not defined pointwise). We could additionally have almost everywhere pointwise convergence, but this requires proof: it is not direct from Hilbert space rules. Answer. As $f_0 = 0$, there is no constant ...

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If $\lambda=0$ then clearly $f$ must be zero, so let us assume $\lambda\neq 0$. Then given any function $f_0:[0,2)\to \mathbb{C}$, there is a unique function $f:\mathbb{R}\to\mathbb{C}$ extending $f_0$ and satisfying the functional equation. Let us take $f_0$ to be the constant function $1$, and let $f$ be the unique extension satisfying the functional ...

2

It is not necessarily true. Consider $x_n = 1/\sqrt n \not\in \ell_2$. Then for any $y_n \in \ell_2$ $$\| x_ny_n \|_2 = \sqrt{\sum_{i=1}^\infty |x_ny_n|^2} \leq \sqrt{\sum_{i=1}^\infty |y_n|^2} = \| y_n \|_2$$ That is, $x_ny_n \in \ell_2$ for all $y_n \in \ell_2$.

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You are right to be skeptical; your limit $a$ does not work. Hint: Knowing that $$\sup_k |a_{n,k} - a_{m,k}| → 0$$ means that for each $k$, $a_{n,k}∈ℝ$ is a ... Proof that the map $\phi: \ell^1 → c_0^*$ given by $T(x)(a) = \sum a_i \overline{x_i}$ is an isometric isomorphism.

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Let $X\in M$. Then, for any $Y\in M$ you have $XY\omega\in M\omega$. So $XP=PXP$. If you do this for $X$ selfadjoint and take adjoints on the equality, you get $XP=PX$. As selfadjoints span $M$, you get that $P\in M'$.

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What does $\frac{R}{z}$ mean? It means that someone is a fan of sloppy notation. It means $\frac{1}{z}R$, which makes sense because $\frac{1}{z}$ is a scalar and $R$ is a vector in the vector space of linear operators. Similarly $\frac{z}{R}$ no doubt stands for $z R^{-1}$, which makes sense in Mathematics. The operator $R$ and its inverse are isometric, ...

1

Take $T =\rm{id}$ and $S= \rm{diag}(1,-1)$ on $\Bbb{R}^2$.

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The operator $L$ maps $\{ a_n \}_{n=-\infty}^{\infty}$ to $\{ a_{n+1} \}_{n=-\infty}^{\infty}$. If $e_n$ is the standard basis element defined as a sequence of all $0$'s except for a $1$ in the $n$-th place, then $Le_n = e_{n-1}$, which is why $L$ would be considered to be a left shift. And this follows from the definition $L\{ a_n \} = \{ a_{n+1} \}$: the ...

2

More elementary way: a. If $\lvert\,\lambda\rvert>1$, then $\lambda\not\in\sigma(\mathcal L)$. This is clear as the series $(\lambda-\mathcal L)^{-1}=\lambda^{-1}\sum_{n=0}^\infty (\lambda^{-1}\mathcal L)^n$ converges. b. If $\lvert\,\lambda\rvert<1$, then $\lambda\not\in\sigma(\mathcal L)$. This is clear as the series $(\lambda-\mathcal ... 3 Answer.$S=\{z\in\mathbb C: \lvert\,z\rvert=1\}$. Explanation. The space$\ell^2(\mathbb Z)$is isometric to$L^2(\mathbb T)$, where$\mathbb T$is the unit circle (equivalently, the domain of$2\pi-$periodic functions), and this is done by the transfomation $$\varPhi\big((a_k)_{k\in\mathbb Z}\big)=\sum_{k\in\mathbb Z}a_k\mathrm{e}^{ikx},$$ as ... 1 HINT.If$a_{n+1}-z a_n=b_n$then$a_2=z a_1+b_1$,$a_3=z a_2+b_2=z^2 a_1+z b_1+b_2$,$a_4=z a_3+b_3=z^3a_1+z^2b_1+z b_2+b_3$, etc. If$|z|=1$and$b_n= z^{n}/(1+|n|)$, what happens to$|a_n-z^{n-1}a_1|$for large positive$n$? 1 It is sufficient to check things for$\mathcal{H}_{K,0}$. Let's look at the Fourier coefficients of the function $$f=\sum_{i=1}^{m}a_{i}K_{u_{i}}.$$ By definition $$K(u,v)=\sum_{h\in\mathbb{Z}^{s}}w(h)e^{2\pi ih^{T}(u-v)},$$ and so $$f(v)=\sum_{i=1}^{m}a_{i}\sum_{h\in\mathbb{Z}^{s}}w(h)e^{2\pi ... 2 Counterexample: f_n = g_n any sequence converging weakly but not strongly. 0 It is just a weighted L^p-space. The norm is given by$$ \|u\|_{L^2_\rho} ^p:=\int_D |u(x)|^2 \rho(x) dx, $$scalar product is$$ (u,v)_\rho = \int_D \rho(x) u(x)v(x)dx. $$2 No, it's not true. Let$$ \psi_i^n = \begin{cases} e_i-\frac{1}{n}\sum_{j=1}^n e_j,\quad &1\le i\le n \\ e_i,\quad & i>n \end{cases} $$Then \|\psi_i^n-e_i\|\le 1/n for all i. On the other hand,$$ \sum_{i=1}^n \psi_i^n =0 $$and the span of \{\psi_i^n\} is orthogonal to the vector \sum_{i=1}^n e_i. 1 Suppose \lambda is an isolated point of the spectrum of T. Let F be a continuous function that is identically 1 near \lambda, and is 0 on the remaining part of the spectrum \sigma(T)\setminus\{\lambda\}. Then F is either 1 or 0 on the spectrum, which makes P=F(T) an orthogonal projection. Clearly (T-\lambda I)P=0 because ... 0 I think I solved it just changing here https://en.wikipedia.org/wiki/Discontinuous_linear_map in General Existence Theorem with T(e_n,e_n)=n and T(e_n,e_m)=0 when n different m and {e_i} is an orthonormal basis. 1 Using the cyclic property of trace and linearity of expectation/trace we have: \begin{eqnarray} \langle y, A X \rangle &=& E [ y^T X b ] \\ &=& E [ \operatorname{tr} ( y^T X b )] \\ &=& E [ \operatorname{tr} ( b y^T X )] \\ &=& \operatorname{tr} ( E [ b y^T ] X ) \\ &=& \operatorname{tr} (E[y b^T]^T X ) \\ &=& ... 0 You are looking at the following eigenvalue problem on [0,\infty):$$ Lf= - \frac{d}{dx}\frac{1}{1+x}\frac{df}{dx}+(x+1)f=\lambda\frac{1}{x+1}f $$If you can explicitly solve for any particular \lambda, then you can determine if the equation is in the limit point or limit circle case at \infty. If you are in the limit circle case, then all ... 1 No, there is not printing error, First of all you should consider any subspace of a Hilbert space like "H" is a Hilbert space and any Hilbert space has a unique decomposition like: H=M+M^{\perp}. Then according to uniqueness of decomposition, M_1 and K are decomposition of Hilbert space M. And N_1 and K are decomposition of Hilbert space ... 5 No, the book doesn't have any typo. First let me give you an example to have an imagination of what is going on. The simplest example to imagine is the following. Let M_1 be the x-axis in \mathbb{R}^3, N_1 be the y-axis, K be the z-axis. Obviously M_1, N_1 and K are mutually disjoint subspaces of \mathbb{R}^3, now you can see that M ... 0 We know that in a Hilbert space S^\perp is always a closed subspace for any subset S of \mathcal H. It is very easy to see that$$M=\{e_{2i-1};i\geq 1\}^\perp$$2 Your idea is not wrong, it just needs to be written down more carefully. I think there are a few points that you are slightly misunderstanding. To prove that your subspace M is closed we need to prove that every convergent sequence in M converges to a point of M. So, as you do, let's take a sequence X_k = \left((x_n^k)_{n \in \mathbb N}\right)^{k ... 0 If X_{k} = \bigl((x_{n}^{k})_{n \in \mathbb{N}}\bigr)^{k \in \mathbb{N}} converges to X= (x_{n})_{n \in \mathbb{N}} in \ell^2, then \lim_{k\to\infty}x_n^k=x_n for all n. Another way is the following. Let \phi_n\colon\ell^2\to\mathbb{R} be the linear functional defined by \phi_n\bigl((x_{n})_{n \in \mathbb{N}}\bigr)=x_n. \phi_n is continuous ... 3 For part one, Gram-Schmidt is indeed the way to go. Let f_1 = e_1 + e_2, f_2 = e_3 + e_4, f_3 = e_2 + e_3. Then, since f_1 and f_2 are already orthogonal to each other, you need only to normalize them. Since e_1 and e_2 are orthogonal, you have ||f_1||^2 = ||e_1||^2 + ||e_2||^2 = 2 so ||f_1|| = \sqrt{2} and u_1 = \frac{e_1 + ... 2 You know that \{e_n\}_{n\in \mathbb{N}} is orthonormal. So let (a,b) be the notation for inner product of the space. Gram-Schmidt is the way to go actually,as you guessed. Convince yourself that with Gram-Schmidt you find$$ f_1 = \frac{e_1+e_2}{\sqrt{2}}, \quad f_2=\frac{e_3+e_4}{\sqrt{2}}, \quad f_3=\frac{e_1-e_2-e_3+e_4}{2}$$The last vector is ... 2 Your idea of using Graham-Schmidt for determining the orthogonal basis is absolutely right, but the answer you present has gone wrong. All you know about the$e_i$is that for all$i$,$e_i \cdot e_i = 1$and for all$i \neq j$,$e_i \cdot e_j = 0$. But that is quite a lot to know, and enough to do G.S. Start from basis element$b_1 =k e_1$; normalizing ... 0 As mentioned in the comments, your use of the term "orthonormal basis" is unconventional : Usually a complete orthonormal set is an orthonormal basis. However, if the space is infinite dimensional, then such a set cannot be a Hamel basis : Any orthonormal set is linearly independent: You don't need uncountable sums here! If$A$is an orthonormal set and ... 2 Let$V$be the space of functions spanned by$f$and$g$; that is, the space of all linear combinations$a f + b g$(that is, the function$x \mapsto a f(x) + b g(x)$) where$a,b$are real numbers. Then, assuming$f$and$g$are linearly independent,$V$is a two-dimensional inner product space — you really can think of this in the same fashion as any ... 2 Take a partition of$[a,b]$into$n$equally-spaced subintervals. Approximate a function by assigning to each subinterval its value at the middle of that subinterval, resulting in a vector in$\mathbb{R}^n$. To check whether two functions are orthogonal, you simply take their inner product in$\mathbb{R}^n\$. That is, you multiply the functions on the ...

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